Calculating Short Circuit of Parallel Generators

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02-09-12 03:49 PM

Join Date:Location:Posts:

Aug 2010Lambertville, MI

4

Calculating short circuit of parallel generators

I have (6) 1.25MVA generators feeding a paralleling switchgear. To determine theswitchgear short circuit rating, do I calculate the short circuit of each individualgenerator and multiply by (6) generators? Or do I consider the system as (1)7.5MVA generator with an equivalent parallel Xd" (1/Xd" total = 1/Xd"1 + 1/Xd"2+ ...)?

#1

Junior MemberNoooorm

02-09-12 04:07 PM


Feb 2003Wisconsin

7,465

Good idea.

Actually I put six of them into my software (PTW by SKM) and let it do itss thing.

#2

Moderatorjim dungar

Originally Posted by Noooorm

I have (6) 1.25MVA generators feeding a paralleling switchgear. To determine theswitchgear short circuit rating, do I calculate the short circuit of each individualgenerator and multiply by (6) generators?...

Forum

Calculating short circuit of parallel generators http://forums.mikeholt.com/showthread.php?t=143070

1 of 6 11/17/2012 6:56 PM

Just because you can, doesn't mean you should.

02-09-12 04:25 PM



4

Could you? I don't have the software. They're 400V, 50hZ. I guess go with theworst case, 0.09 for X"d. I'll see which answer it's closest to.

#3


02-09-12 04:29 PM


May 2005Redmond, WA

694

#4

Senior Memberrcwilson


2 of 6 11/17/2012 6:56 PM

Calculate for one generator and multiply by 6.

I short circuit = FLA/Xd" = [1.25 x 1000/(.480 x 1.732) ] /0.2 = 1504 /0.2 A =7517 Amps per generator.7517 Amps x 6 = 45,106 Amps

Or using the MVA method:Assume typical Xd" = 0.2, Short Circuit MVAsc = 1.25 MVA/ 0.2 = 6.25 MVA,I short circuit = MVAsc x 1000 / (kV x1.732) = 7517 amps (assuming 480 V).Total short circuit = 7517 amps x 6 = 45,106 Amps.

Same answer if we multiply MVAsc of one generator by 6 and calculate amps.6.25 x 6 = 37.5 MVA total short circuit power,I = 37.5x 1000/(0.480 x 1.732) = 45,107 Amps.

Since the impedance Xd" is per unit or per cent and not in actual ohms we don'tneed to calculate the parallel impedance. Example:

Six each 1.25 MVA units = 7.5 MVA all with Xd"=0.2. MVAsc = 7.5/0.2 = 37.5 MVA,same answer as above.

Bob Wilson P.E.

Originally Posted by Noooorm

I have (6) 1.25MVA generators feeding a paralleling switchgear. .... calculate theshort circuit of each individual generator and multiply by (6) generators? Or do Iconsider the system as (1) 7.5MVA generator with an equivalent parallel Xd" (1/Xd"total = 1/Xd"1 + 1/Xd"2 + ...)?

02-09-12 04:31 PM


Feb 2003Seattle, WA

16,105

I think you will get the same answer either way. If the generators were notidentical, however, a method closer to your second one would be needed. But

my preference is to use SKM, and leave the messy math to someone else.

Charles E. Beck, P.E., SeattleComments based on 2008 NEC unless otherwise noted.

#5

Moderatorcharlie b

02-09-12 04:33 PM #6


3 of 6 11/17/2012 6:56 PM


May 2005Redmond, WA

694

400V will change you short circuit current to 480/400 x 45,107 = 54, 128 if yourgenerators have the typical 0.2 Xd". I think most generators in this size have Xd" =0.12 - 0.25 per unit ( 12-25%).

Don't forget to add in any contribuiton form motor loads.

Bob Wilson P.E.


02-09-12 04:48 PM



4

Thank you! You are gentlemen and scholars.

#7


02-10-12 02:26 PM

Join Date:Posts:

Jun 20035,229

That was my first though too, but it doesn't seem to work. Consider (2) 1KWgenerators, z = 0.2. That's 5000 amps SC per genrators, or 10000 amps total.

Using the second method, you would have a 2 KW generator, and z =0.2 in parallelwith 0.2 would be 0.1. 2000/ 0.1 = 20,000 amps. Unless my math is wrong, thatanswer is too high.

For the second generator, we are both doubling the capacity, and halving theimpedence, giving a 4 fold increase, when the real fault current is only going up bya factor of 2.

#8

Senior Membersteve66

Originally Posted by charlie b

I think you will get the same answer either way. If the generators werenot identical, however, a method closer to your second one would beneeded. But my preference is to use SKM, and leave the messy math

to someone else.

#9


4 of 6 11/17/2012 6:56 PM

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02-10-12 05:15 PM


May 2005Redmond, WA

694

The generator impedance is expressed in percent or per unit. It is the measured orcalculated impedance based on the generator nameplate data. That number is notthe actual real world ohms that we could measure with an ohmmeter (if there wassuch a thing as an AC ohmmeter).

If 0.2 was real world ohms, then adding them like parallel resistances is correct.

But since it is a percentage or per unit number we don't change it when we areparalleling identical units.

Said another way, the per unit impedance of two paralleled 1 MVA generators withimpedance = 0.2 pu is still 0.2 per unit. The math works because on individualunits, it's 0.2 per unit of 1 MVA and on two parallel units it's 0.2 of 2 MVA.

If you want to do it in real world ohms just convert to ohms:

MVA = kV^2 / Z. (Power = voltage squared/ ohms impedance). or Z= kV^2 /MVA.

Zgen = (0.400kV x 0.400kV)/ 1.25 MVA = 0.1280 ohms. This is the 100% or 1.0per unit impedance. If Xd"= 0.2 pu, then Zgen = 0.2 x 0.1280= 0.0256 ohms

Short circuit current for one 1.25 MVA generator at the OP's 400V is limited only bythe generator impedance of 0.0256 ohms. Isc = 400V / (0.0256 x 1.732) = 9,021amps. Six generators would be 6 x 9021= 54,126 Amps.

With six generators in parallel the impedance for a short circuit would be theparallel impedance of six generators = 1/ (1/0.0256 + 1/0.0256 + 1/0.0256 +1/0.0256 + 1/0.0256 + 1/0.0256) = 0.0256/6

Z= 0.00427. Isc = V/Z where V= phase- neutral volts = 400 /(1.732 x .00427) =54,126 amps.

I probably confused some of us, but I was trying to explain the per unit, per centconcept.

Bob Wilson P.E.


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Calculating Short Circuit of Parallel Generators

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